Question

The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point C are (0, −3). The origin is the midpoint of the base. Find the coordinates of the points A and B. Also, find the coordinates of another point D such that ABCD is a rhombus.

Answer 1

Given:

ΔABC is equilateral

Base BC lies on the y–axis

C = (0, –3)

Origin (0, 0) is midpoint of BC

B = (0, b), C = (0, −3)

`0 =  (b+(-3))/2`

b − 3 = 0

b = 3

B = (0, 3)

C = (0, –3)

In an equilateral triangle,

height = `sqrt3/2 xx "side"`

Height = `sqrt3/2 xx 6`

`= 3sqrt3`

Base lies on y-axis, so the altitude is horizontal, meaning A lies on the x-axis but at distance `3sqrt3` from O.

`A = (+-3sqrt3, 0)`

A₁ = `(3sqrt3, 0)`

A₂ = (`-3sqrt3, 0`)

A rhombus has all sides equal, and ABCD is cyclic order.

A = (`3sqrt3`​, 0), B = (0, 3), C = (0, −3)

To form a rhombus ABCD, the 4th point D is:

D = A + C − B

D = (`3sqrt3`​, 0) + (0, −3) − (0, 3)

D = (`3sqrt3​`, −6)

Base points:

B = (0, 3), C = (0, −3)

Vertices of equilateral triangle:

A = (`3sqrt3​`, 0) or A = (`−3sqrt3​`, 0)

Point D such that ABCD is a rhombus:

If A = (`3sqrt3`, 0): 

D = (`3sqrt3​`, −6)

If A = (`−3sqrt3`, 0):

D = (`-3sqrt3​`, −6)